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You want to store \(165 \mathrm{g}\) of \(\mathrm{CO}_{2}\) gas in a \(12.5-\mathrm{L}\) tank at room temperature \(\left(25^{\circ} \mathrm{C}\right) .\) Calculate the pressure the gas would have using (a) the ideal gas law and (b) the van der Waals equation. (For \(\mathrm{CO}_{2}\) \(\left.a=3.59 \text { atm } \cdot \mathrm{L}^{2} / \mathrm{mol}^{2} \text { and } b=0.0427 \mathrm{L} / \mathrm{mol} .\right)\)

Short Answer

Expert verified
Pressure is 7.33 atm (Ideal Gas Law) and 7.19 atm (van der Waals).

Step by step solution

01

Convert Mass to Moles for CO2

First, determine the number of moles of carbon dioxide (\(\mathrm{CO}_2\)). The molar mass of \(\mathrm{CO}_2\) is approximately 44.01 g/mol. Use the formula: \[\text{moles of } \mathrm{CO}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{165 \text{ g}}{44.01 \text{ g/mol}} \approx 3.75 \text{ moles}\]
02

Use Ideal Gas Law to Find Pressure

Apply the ideal gas law \(PV = nRT\) to calculate pressure. Here, \(R\) is the ideal gas constant \(0.0821 \, \text{atm} \, \cdot \, \text{L/mol} \, \cdot \, K\), \(T\) is the temperature in Kelvin (\(T = 25 + 273 = 298\, \text{K}\)), \(n\) is the number of moles (\(3.75\)), and \(V\) is the volume (\(12.5\, \text{L}\)).\[P = \frac{nRT}{V} = \frac{3.75 \, \text{mol} \times 0.0821 \, \text{atm} \, \cdot \, \text{L/mol} \, \cdot \, K \times 298 \, \text{K}}{12.5 \, \text{L}} \approx 7.33 \, \text{atm}\]
03

Calculate Pressure Using van der Waals Equation

The van der Waals equation is \[\left(P + \frac{an^2}{V^2}\right)(V-nb) = nRT\] where \(a = 3.59 \, \text{atm} \, \cdot \, \text{L}^2/\text{mol}^2\) and \(b = 0.0427 \, \text{L}/\text{mol}\).First, calculate: \(n^2/V^2\) and \(bn\):\[\frac{n^2}{V^2} = \frac{(3.75)^2}{(12.5)^2} = 0.09\]\[nb = 3.75 \times 0.0427 = 0.16 \, \text{L}\]Substitute these into the van der Waals equation:\[\left(P + \frac{3.59 \times 0.09}{12.5^2} \right)(12.5 - 0.16) = 3.75 \times 0.0821 \times 298\]Solve for \(P\):\[P = \frac{3.75 \times 0.0821 \times 298}{12.34} - 0.036 \approx 7.19 \, \text{atm}\]
04

Compare Results

Summarize the results obtained from both calculations:1. Using the Ideal Gas Law, the calculated pressure is approximately \(7.33 \text{ atm}\).2. Using the van der Waals equation, the calculated pressure is approximately \(7.19 \text{ atm}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas Law
The Ideal Gas Law is a fundamental equation used to describe the behavior of ideal gases. It is expressed as \( PV = nRT \). In this equation:
  • \( P \) represents the pressure of the gas
  • \( V \) stands for volume
  • \( n \) is the number of moles of the gas present
  • \( R \) is the ideal gas constant
  • \( T \) denotes the temperature in Kelvin
This equation assumes that gas molecules do not attract or repel each other and that they occupy no space, which greatly simplifies calculations for ideal conditions.
To use the equation effectively, convert given measurements like temperature to Kelvin and calculate the moles of gas if the mass is provided. In real-life situations, however, gases often deviate from ideal behavior under high pressure or low temperature conditions.
Van der Waals Equation
The Van der Waals Equation is a modified version of the Ideal Gas Law. It accounts for the actual volume occupied by gas particles and the attraction between them. This formula is especially useful for real gases, which deviate from the ideal model. The equation is given by: \[\left( P + \frac{an^2}{V^2} \right)(V-nb) = nRT\]
  • The term \(a\) corrects for intermolecular forces, and \(b\) corrects for the volume occupied by the gas molecules.
  • Values for \(a\) and \(b\) are specific to each type of gas.
Let's break it down: while \(P\), \(V\), \(n\), \(R\), and \(T\) remain the same as in the Ideal Gas Law, the terms \( \frac{an^2}{V^2} \) and \( nb \) are adjustments that lead to more accurate predictions. Thus, this formula provides a more precise pressure value when gas molecules have significant interactions or occupy notable space in the container.
Molar Mass
Molar Mass is a measure of the mass of one mole of a given substance, typically expressed in grams per mole (g/mol). This concept is crucial to connect macroscopic measurements (like mass) with microscopic quantities (like moles) in chemistry.
For carbon dioxide (\( \text{CO}_2 \)), the molar mass is calculated by adding the atomic masses of one carbon atom and two oxygen atoms, which are approximately 12.01 g/mol and 16.00 g/mol each, respectively. Therefore, the molar mass of \( \text{CO}_2 \) is approximately 44.01 g/mol.
  • To find the number of moles, divide the mass by the molar mass using the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \).
Understanding molar mass allows us to convert between mass and moles, facilitating calculations involving other gas laws.
Pressure Calculation
Pressure Calculation involves determining the force exerted by gas particles on the walls of their container. Both the Ideal Gas Law and the Van der Waals Equation can be used for calculating pressure but have different application scenarios.
When using the Ideal Gas Law, the calculation simplifies to: \[ P = \frac{nRT}{V} \]In this scenario, the gas is assumed to behave ideally, with negligible volume occupied by particles and no intermolecular forces. For high accuracy under non-ideal conditions, use the Van der Waals Equation:\[\left( P + \frac{an^2}{V^2} \right)(V-nb) = nRT\]This considers real-world behaviors by correcting factors of molecular attraction and particle volume.
Deciding which formula to use largely depends on the conditions and the degree of accuracy needed for the pressure being calculated. The Ideal Gas Law provides ease of calculation, whereas Van der Waals gives more precise results when exact conditions matter.

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Most popular questions from this chapter

A flask is first evacuated so that it contains no gas at all. Then, \(2.2 \mathrm{g}\) of \(\mathrm{CO}_{2}\) is introduced into the flask. On warming to \(22^{\circ} \mathrm{C},\) the gas exerts a pressure of \(318 \mathrm{mm}\) Hg. What is the volume of the flask?

Nitrogen monoxide reacts with oxygen to give nitrogen dioxide. $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) $$ (a) You wish to react \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) in the correct stoichiometric ratio. The sample of NO has a volume of 150 mL. What volume of \(\mathrm{O}_{2}\) is required (at the same pressure and temperature)? (b) What volume of \(\mathrm{NO}_{2}\) (at the same pressure and temperature) is formed in this reaction?

What is the total pressure in atmospheres of a gas mixture that contains \(1.0 \mathrm{g}\) of \(\mathrm{H}_{2}\) and \(8.0 \mathrm{g}\) of \(\mathrm{Ar}\) in a \(3.0-\mathrm{L}\) container at \(27^{\circ} \mathrm{C}\) ? What are the partial pressures of the two gases?

Nitrogen monoxide reacts with oxygen to give nitrogen dioxide: $$ 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2}(\mathrm{g}) $$ (a) Place the three gases in order of increasing rms speed at \(298 \mathrm{K}\) (b) If you mix \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) in the correct stoichiometric ratio and NO has a partial pressure of \(150 \mathrm{mm} \mathrm{Hg},\) what is the partial pressure of \(\mathrm{O}_{2} ?\) (c) After reaction between \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) is complete, what is the pressure of \(\mathrm{NO}_{2}\) if the NO originally had a pressure of \(150 \mathrm{mm} \mathrm{Hg}\) and \(\mathrm{O}_{2}\) was added in the correct stoichiometric amount?

Calculate the rms speed for CO molecules at \(25^{\circ} \mathrm{C} .\) What is the ratio of this speed to that of \(\mathrm{Ar}\) atoms at the same temperature?

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